Analytical Mechanics This page intentionally left blank
Download 10.87 Mb. Pdf ko'rish
|
|
· HQ
2 + a X H a 2 − X 2 dX,
and HP = (b/a)HQ. Therefore area (AF P ) = b/aarea (AF Q). (5.80) By Kepler’s second law it follows that the area (AF P ) = L z t/2m, yielding L z
2m = b a area (AF Q) = b a
− area (QF C)] = b a a 2 ξ 2 − CF · QH
2 = ab 2 (ξ − e sin ξ). This finally leads to Kepler’s equation (5.75). The solution of Kepler’s equation is given by the following theorem (here, to avoid any confusion with number e = 2.718 . . ., the eccentricity is denoted by . See also Problem 6). T heorem 5.4 The eccentric anomaly is an analytic function of the eccentricity for | | < 1/e and of the mean anomaly l. Its series expansion is given by ξ( , l) = ∞ m =0 c m (l) m m! , (5.81)
196 Motion in a central field 5.4 where
c 0 (l) = l, (5.82) c m (l) = d dl m −1 (sin l) m for all m ≥ 1. (5.83)
The series (5.81) converges uniformly in l and for all values of the eccentricity | | < 1/e. Remark 5.5 Newton proved that the solution of Kepler’s equation, which expresses the coordinates (x, y) of the point particle along the orbit as a function of time, cannot be an algebraic function. His proof can be found in the book by Arnol’d (1990, Chapter 5). The proof of Theorem 5.4 is a consequence of Lagrange’s formula, to be discussed in the next section. As an exercise we now verify that the first two terms in the series expansion (5.71) are correct. Since when the eccentricity is zero we have that ξ = l, it is natural to seek the solution of Kepler’s equation in the form of a series in with coefficients depending on the mean anomaly l and with the zero-order term equal to l: ξ( , l) = l + ξ 1 (l) +
2 ξ 2 (l) + O( 3 ). (5.84)
Substituting equation (5.84) into (5.75) we find l + ξ
1 + 2 ξ 2 − [sin l cos( ξ 1 + 2 ξ 2 ) + cos l sin( ξ 1 + 2 ξ 2 )] = l + O( 3 ). Using the series expansion of the sine and cosine functions, up to second order, this yields ξ 1
2 ξ 2 − sin l − 2 ξ 1 cos l =
O( 3 ). From the latter expression, by equating powers of , it follows that ξ 1 (l) = sin l, ξ 2 (l) = ξ 1 (l) cos l = sin l cos l, and hence ξ( , l) = l + sin l + 2 sin l cos l + O( 3 ), which is in agreement with (5.83). The proof of the uniform convergence, for | | < e −1
(5.81) of the solution of Kepler equation follows from the formula (see Wintner 1941)
c m (l) = d m −1 dl m −1 (sin l) m = [m/2] k =0 (m − 2k)
m −1 ( −1) k 2 m −1 m k sin[(m
− 2k)l], 5.5 Motion in a central field 197 where [m/2] denotes the integer part of m/2. This formula can be deduced from (sin l) m = m k =0 m k 1 (2i) m ( −1) k e i(m−2k)l . Applying Stirling’s formula (Dieudonn´ e 1968, p. 130) m! ∼ √ 2πm
m +1/2
e −m (for m → +∞), since
| sin[(m − 2k)l]| ≤ 1 and (m − 2k) m −1 ≤ m m −1 , we find max
0≤l≤2π |c m (l) | ≤
1 2 m −1 m m −1 [m/2]
k =0 m k ≤ 1 2 m −1 m m −1 2 m = 2m m −1 , yielding max
0≤l≤2π m m! c m (l) ∼ 2 m m m −1 √ 2πe
−m m m +1/2 = 2 π e m m −3/2 m
. It follows that |ξ( , l)| ≤ 2 π ∞ m =0 (e ) m
∞ if 0
≤ < 1/e, i.e. the radius of convergence of the power series for the eccentricity ∞ m =0 max
0≤l≤2π |c m (l) | m m! is at least 1/e. 5.5 The Lagrange formula In the previous section we studied the Kepler equation (5.75) and the convergence of its series solution (5.81), (5.83). The expansion of the eccentric anomaly in power series in terms of the eccentricity is a particular case of a more general formula. This formula was derived by Lagrange in the study of the Kepler equation. In this section we now prove that the series (5.81) is indeed a formal solution of (5.75). Consider the more general problem of the determination of the solution x = X(y, ) of an equation x = y + εf (x), (5.85)
where is a parameter and f is an analytic function of x such that f (0) = 0. If |εf (x)| < 1 the implicit function theorem yields the existence of a unique solution
198 Motion in a central field 5.5 x = X(y, ) in a neighbourhood of x = y. We can seek the series expansion of X as a power series in ε with coefficients depending on y: X(y, ) = y + εX 1 (y) + ε
2 X 2 (y) + · · · .
The Lagrange formula (also called the series inversion, as it yields the inversion of the relation between x and y through a series expansion) is a formula for the explicit determination of the coefficients X n of this expansion, in terms of the function f and of its derivatives. T heorem 5.5 (Lagrange) The solution x = X(y, ) of (5.85) is given by the series of functions (if it converges) X(y, ) = y + ∞ n
ε n X n (y),
(5.86) where
X n (y) = 1 n! d n −1 dy n −1 [(f (y)) n ], (5.87) with the convention (d 0 /dy 0 )f (y) = f (y). In addition, if g is an analytic function such that g(0) = 0, then g(X) = g(y) + ∞ n
ε n G n (y),
(5.88) where
G n (y) = 1 n! d n −1 dy n −1 [(f (y)) n g (y)].
(5.89) The proof is given below. Remark 5.6 The formulae (5.86) and (5.87) are obtained from equations (5.88) and (5.89) by setting g(y) = y. The previous theorem yields the following corollary. C orollary 5.1 The series (5.81), (5.83) solves Kepler’s equation (5.75). Proof It is sufficient to set x = ξ, y = l, ε = e, f (ξ) = sin ξ and to apply the result of Theorem 5.5.
5.5 Motion in a central field 199 Proof of Theorem 5.5 Let x = X(y, ) be the solution; its existence is guaranteed by the implicit function theorem. By differentiating (5.85) we find dX = dy + f (X) dX + f (X) d , (5.90)
from which it follows that ∂X ∂ = f (X)
1 − f (X)
= f (X) ∂X ∂y . (5.91)
Let F and G be any two regular functions. From (5.91) it follows that ∂ ∂ F (X(y, )) = f (X(y, )) ∂ ∂y F (X(y, )), (5.92)
and hence ∂ ∂ F (X(y, )) ∂ ∂y G(X(y, )) = ∂ ∂y F (X(y, )) ∂ ∂ G(X(y, )) (5.93)
as ∂ ∂ F (X) ∂ ∂y G(X) = f (X)
∂F ∂y (X) ∂G ∂y (X) + F (X) ∂ 2 G ∂ ∂y (X),
∂ ∂y F (X) ∂ ∂ G(X) = ∂F ∂y (X)f (X) ∂G ∂y (X) + F (X) ∂ 2 G ∂y∂
(X). From equations (5.92) and (5.93) we deduce by recurrence that for every integer n ≥ 1 and for any analytic function g such that g(0) = 0 we have ∂ n g ∂ n (X(y, )) = ∂ n −1 ∂y n −1 (f (X(y, ))) n ∂g ∂y (X(y, )) . Consequently Taylor’s formula yields g(X(y, )) = g(X(y, 0)) + ∞ n
n n! ∂ n −1 ∂y n −1 f (X(y, 0)) n ∂g ∂y (X(y, 0)) . Since X(y, 0) = y we find the expression (5.89). Example 5.3 Consider the equation x = y + εx 2 . (5.94) 200 Motion in a central field 5.6 Applying the Lagrange formula we find the solution X(y, ) = y + ∞ n =1 ε n n! (2n)!
(n + 1)! y n +1 . On the other hand, the solution of (5.94) such that X(0, ) = 0 is given by X(y, ) = 1 2ε − 1 2ε 1 − 4εy.
Verify as an exercise, using the Taylor series expansion of √ 1 − 4εy, that the Lagrange series is indeed the correct solution (hint: first show by induction that 2 n
−1)!! = (2n)!/n!). For fixed y, what is the radius of convergence of the series?
5.6 The two-body problem Consider two bodies of mass m 1 and m 2 described by the position vectors r 1 and r
2 , respectively, in R 3 . Assume the two bodies interact through a central potential V ( |r 1 − r 2 |). We can prove the following. T heorem 5.6 Let r = r 1
2 , r M = m 1 r 1 + m 2 r 2 m 1 + m 2 (5.95) be, respectively, the relative position vector of the two bodies and the position vector of the centre of mass of the system. Then the acceleration of the centre of mass with respect to an inertial system is zero, and in a reference system having the centre of mass as its origin the equations of motion are given by m¨ r =
−∇ r V, (5.96) where
m = m 1 m 2 m 1 + m
2 (5.97)
is the so-called reduced mass of the two-body system. In addition, since r 1 = r M + m 2 m 1 + m
2 r, r 2 = r
M − m 1 m 1 + m 2 r, (5.98) the trajectories of the two points are planar curves lying in the same plane, and similar to each other, with similarity ratio m 1 /m 2 .
5.7 Motion in a central field 201 Proof
The verification that ¨ r M = 0 is left to the reader. In addition, m¨ r = m¨ r 1 − m¨r 2 = m 1 m 2 m 1 + m 2 − 1 m 1 ∇ r 1 V + 1 m 2 ∇ r 2 V = −∇ r V. The remaining claims are of immediate verification. We saw in Section 4.9 that because of the (rigid) translation invariance of the two-body system, due to the absence of forces external to the system, the centre of mass follows a linear uniform motion. The initial value problem with six degrees of freedom is thus reduced to the problem of the motion of a point particle, with mass equal to the reduced mass of the system, in a central potential field. For this problem, the considerations of the previous sections apply. 5.7 The
n-body problem The study of the problem of n bodies interacting through a Newtonian potential is central to the study of celestial mechanics (for a very readable introduction, see Saari 1990). In the previous section we saw how the two-body problem is integrable and can be reduced to the motion of a single point in a central potential field. If n ≥ 3, the resulting motion is much more complicated, and in this short introduction we only list the most classical and elementary results. For a more detailed exposition, we recommend Wintner (1941), Siegel and Moser (1971), Alekseev (1981) and the monographs of Pollard (1966, 1976) which inspired this section. Consider n bodies with masses m i , i = 1, . . . , n and corresponding position vec- tors r i (measured in an inertial reference frame). Let G indicate the gravitational constant. The force of the gravitational attraction between any pair (i, j) has the direction of r j − r
i and intensity equal to Gm i m
/r 2 ij , where r ij = |r i − r j |. The
equations describing the motion of the bodies are of the form m i ¨ r i = n j =1,j=i Gm i m j r 2 ij r j − r
i r ij , (5.99)
where i = 1, . . . , n. Remark 5.7 The problem of the existence of solutions, local in time, of the initial value problem for equation (5.99), for prescribed initial conditions r i (0), v
i (0), and of the possibility of their continuation has been widely studied. Here we only note that from the theorem of the existence and uniqueness for ordinary differential equations there follows the existence of a solution of the system (5.99) for a sufficiently small time interval, assuming that at time t = 0 the relative distances |r i
− r j (0) | of the points are bounded from below by a constant r 0 > 0. 202 Motion in a central field 5.7 The system of differential equations (5.99) is a system of order 6n. Denoting by r M = 1 M n i =1 m i r i (5.100)
(where M = n i =1 m i represents the total mass of the system) the position vector of the centre of mass, it is easy to verify, by summing over all i in equations (5.99), that ¨ r M = 0.
(5.101) From this, it follows that the centre of mass moves with a linear uniform motion in the chosen inertial frame of reference. Hence the coordinates and the velocity of the centre of mass constitute a set of six first integrals of the motion for the system. In what follows we suppose that the chosen frame of reference has origin coinciding with the centre of mass, and axes parallel to those of the initial inertial system, so that r M = ˙r M = 0. (5.102) In addition it is easy to verify that in this frame of reference the energy integral can be written as E = T + V = n i
1 2 m i |˙r
i | 2 − 1≤i Gm i m j r ij , (5.103)
where T =
n i =1 1 2 m i |˙r
i | 2 and V =
− 1≤i Gm i m j r ij represent the kinetic and potential energy of the system. An equivalent formulation of the conservation of E can be obtained when considering the polar moment of inertia of the system I =
1 2 n i =1 m i r 2 i , (5.104) where r i = |r i |. Indeed, by differentiating twice with respect to time, we find ¨ I =
i =1 m i |˙r
i | 2 + n i =1 r i · m i ¨ r = 2T + n i =1 r i · m i ¨ r i , 5.7 Motion in a central field 203 and, recalling (5.99), it follows that ¨ I = 2T +
n i =1 n j =1,j = / i Gm i m j r 3 ij [r i · r j − r 2 i ] = 2T + 1 2 n i =1 n j =1,j = / i Gm i m j r 3 ij [r 2 j − r
2 i − r 2 ij ] = 2T − 1 2 n i =1 n j =1,j = / i Gm i m j r ij = 2T + V = E + T. (5.105) The identity ¨ I = E + T is called the Lagrange–Jacobi identity. The coordinates, the velocity of the centre of mass of the system and the total energy E constitute a set of seven independent constants of the motion for the n-body problem. Three more constants are given by the conservation of the total angular momentum L: L =
n i =1 m i r i × ˙r
i . (5.106) Indeed, since ˙r i × ˙r i = 0, we find ˙ L =
n i =1 m i r i × ¨r
i = n i =1 n j =1,j
= / i Gm i m j r 3 ij r i × r j = 0, because in the last summation there appear both r i × r j and r
j × r
i = −r i × r
j with equal coefficients. This yields the proof of the following theorem. T heorem 5.7 The system of equations (5.99) admits ten first integrals of the motion. We end this brief introduction with an important result (see Sundman 1907) regarding the possibility that the system of n points undergoes total collapse, i.e. that all n particles are found in the same position at the same time, colliding with each other (in this case, of course equation (5.99) becomes singular). T heorem 5.8 (Sundman) A necessary condition that the system undergoes total collapse is that the total angular momentum vanishes. Before proving Theorem 5.8, we consider a few lemmas. L emma 5.2 (Sundman inequality) If L is the magnitude of the total angular momentum (5.106) then L 2 ≤ 4I(¨I − E). (5.107)
204 Motion in a central field 5.7 Proof
From definition (5.106) it follows that L ≤ n i =1 m i |r i × ˙r
i | ≤
n i =1 m i r i v i , where v
i = |˙r i |. Applying the Cauchy–Schwarz inequality we find L 2
⎛ ⎝ n i =1 m i r 2 i ⎞ ⎠ ⎛ ⎝ n i =1 m i v 2 i ⎞ ⎠ = 4IT. Hence the result follows from the Lagrange–Jacobi identity (5.105). L emma 5.3 Let f : [a, b] → R be a function of class C 2 such that f (x) ≥ 0, f (x)
≥ 0 for every x ∈ [a, b]. If f(b) = 0 then f (x) ≤ 0 for every x ∈ [a, b]. The proof of Lemma 5.3 is left to the reader as an exercise. L emma 5.4 The polar moment of inertia (5.104) is given by I = 1 2M 1≤i m i m j r 2 ij . (5.108) Proof
We have that n i =1 m i (r i − r j ) 2 = n i =1 m i r 2 i − 2r j · n i =1 m i r i + ⎛ ⎝ n i =1 m i ⎞ ⎠ r
2 j . Hence from (5.102) it follows that n i =1 m i (r i − r j ) 2 = 2 I + Mr
2 j . Multiplying both sides by m j , summing over j and using that (r i − r
j ) 2 = r 2 ij we find n i =1 n j =1 m i m j r 2 ij = 2 I n j =1 m j + M n j =1 m j r 2 j = 4 IM.
From this relation, since r ii = 0, we deduce equation (5.108). Equation (5.108) shows that total collapse implies the vanishing of I (all
particles collide at the origin). Proof of Theorem 5.8 We first show that any total collapse must necessarily happen in finite time, i.e. it is impossible that I(t) → 0 for t → +∞. Indeed, if for t → +∞ we find r ij → 0 for every i and j, then V → −∞. From the Lagrange–Jacobi identity (5.104) it follows that ¨ I → +∞. There then exists a time ˆt such that for every t ≥ ˆt 5.8 Motion in a central field 205 we have ¨ I(t) ≥ 2, and hence I(t) ≥ t 2 + At + B for t → ∞, contradicting the hypothesis I → 0. Hence any total collapse must happen at some finite time t 1 . We have just showed that V → −∞, I → 0 and ¨I → +∞ as t → t 1 , and thus an application of Lemma 5.3 to I(t) yields that ˙I(t) ≤ 0 for t 2
1 . Multiplying both sides of the Sundman inequality (5.107) by −˙I/I we find − 1
L 2 ˙I I ≤ E ˙I − ˙I¨I, and integrating both sides of the latter with respect to time, we find for t ∈ (t
2 , t
1 ) that 1 4 L 2 log
I(t 2 ) I(t) ≤ E[I(t) − I(t 2 )]
1 2 [˙ I 2 (t) − ˙I 2 (t 2 )] ≤ EI(t) + C, where C is a constant. Hence 1 4 L 2 ≤ E I(t) + C
log( I(t
2 )/ I(t)) , which tends to 0 for t → t 1
5.8 Problems
1. Study the existence and stability of circular orbits for the following central potentials: V (r) = ar −3/2
+ br −1 , V (r) = ae br , V (r) = ar sin(br), for varying real parameters a and b. 2. Find a central potential for which the polar angle varies with time as ϕ(t) = arctan(ωt), with ω ∈ R fixed by the initial conditions. 3. Solve the orbit equation for the potential V (r) = −kr −1
−2 , where
k > 0 and a are prescribed constants. (Answer: r = p/(1 + e cos(ωϕ)) with ω 2 = 1 + 2am/L 2 z .) 4. A spherical galaxy has approximately constant density near its centre, while the density tends to zero as the radius increases. The gravitational potential it generates is then proportional to r 2 , a constant for small values of r, and proportional to r −1 for large values of r. An example of such a potential is given by the so-called isochronous potential (see Binney and Tremaine 1987, p. 38): V (r) =
− k b + √ b 2 + r 2 . 206 Motion in a central field 5.8 Introduce an auxiliary variable s = 1 + r 2 /b 2 + 1. Prove that, if s 1 and s
2 correspond, respectively, to the distance of the pericentre and of the apocentre, the radial period is given by T r = 2πb
√ −2E
1 − 1 2 (s 1 + s 2 ) . In addition, prove that s 1 + s 2 = 2(1
− k/2Eb), and hence that T r = 2πk
( −2E)
3/2 ; thus T r depends only on the energy E, and not on the angular momentum L (this is the reason V (r) is called isochronous). Prove also that the increment Φ of the azimuthal angle between two consecutive passages at pericentre and apocentre is given by Φ = π 2 1 + L √ L 2 + 4kb
. Note that Φ → π when b → 0. 5. Find the series expansion of the solution of the equation ξ − e cos ξ = l and prove that it converges uniformly in l for |e| small enough. 6. Let y = sin x = x − x
3 /3! + x
5 /5! +
O(x 7 ). Compute the expansion of x = X(y) up to terms of order O(y
7 ). Verify the accuracy by comparing with the Taylor series expansion of x = arcsin y. 7. Let y = x − x 2
3 − x
4 + · · · . Compute x = x(y) up to fourth order. (Answer: x = y + y 2 + y 3 + y
4 + · · · .) 8. Let y = x − 1/4x
2 + 1/8x
3 − 15/192x 4 +
fourth order. (Answer: x = y + 1/4y 2 + O(y 5 ).) 9. Solve, using the Lagrange formula, the equation x = y + ε sin(hx). For fixed y, for which values of ε does the series converge? 10. Solve, using the Lagrange formula, the equation x = y + εx 3 and discuss the convergence of the series. 5.9 Motion in a central field 207 11. In the Kepler problem, express the polar angle ϕ as a function of the average anomaly l and of the eccentricity e up to terms of order O(e
3 ). (Answer: ϕ = l − 2e sin l + 5/4e 2 sin(2l) + O(e 3
12. With the help of Fig. 5.4 show that r = a(1 − e cos ξ) (remember that CF = ae). 5.9
Additional remarks and bibliographical notes In this chapter we studied central motions and we have seen the most elementary results on the n-body problem. What we proved constitutes a brief and elementary introduction to the study of celestial mechanics. According to Poincar´ e, the final aim of this is ‘to determine if Newton’s law is sufficient to explain all the astronomical phenomena’. 1 Laskar (1992) wrote an excellent introduction to the history of research on the stability of the solar system, from the first studies of Newton and Laplace up to the most recent developments of the numerical simulations which seem to indicate that the motion of the planets, on long time-scales, is better analysed with the tools of the theory of chaotic dynamical systems. This is at odds with the previous firm (but unproven) belief that planets are unchangeable, stable systems. 2 For a deeper study of these topics, we suggest the book of Pollard (1976), of exceptional clarity and depth. In it, after a summary of the fundamental notions of mechanics and analytical mechanics, the reader can find the most important results regarding the n-body problem. The problem with n = 3 is treated in depth, with a careful analysis of all solutions discovered by Lagrange and Euler and of the equilibrium positions of the reduced circular plane problem (where it is assumed that the three bodies belong to the same plane and that two of them—the primary bodies—move with a uniform circular motion around the centre of mass). Another advanced work on the same subject is the book by Meyer and Hall (1992). These authors choose from the first pages a ‘dynamical systems’ approach to celestial mechanics, which requires a greater mathematical background (the present book should contain all necessary prerequisites to such a reading). The monograph by Wintner (1941) is of fundamental importance, but it is not easy to read. In Wintner’s book one can find a discussion of Bertrand’s theorem (1873) (see also Arnol’d 1978a and Albouy 2000). The Kepler equation, and the various analytical and numerical methods for its solution, are discussed in detail by Danby (1988) and by Giorgilli (1990). In 1 ‘Le but final de la m´ ecanique c´ eleste est de r´ esoudre cette grand question de savoir si la loi de Newton explique ` a elle seule tous les ph´ enom`
enes astronomiques’ (Poincar´ e 1892,
p. 1). 2 However, Newton had already expressed doubts in the Principia Mathematica as he considered that it was necessary to allow for the intervention of a superior being in order to maintain the planets near their Keplerian orbits for very long times. 208 Motion in a central field 5.10 the former, one can also find the listings of some easy BASIC programs to solve Kepler’s equation and determine the time dependence in Kepler’s problem. In addition, the book contains a detailed discussion of the problem of determining all elements of a Keplerian orbit (eccentricity, semi-axes, inclination, etc., see Section 9.8) starting from astronomical data. In this book it was impossible, due to space constraints, to go into a more detailed study of the geometric and topological aspects of the two-body and n-body problems (although Appendix 6 partially fills this gap). The articles by Smale (1970a,b) and the discussion by Abraham and Marsden (1978) are excellent but very difficult; the work of Alekseev (1981) is more accessible but less complete. Finally, a curious observation: the Ptolemaic theory of epicycles has recently been interpreted through Fourier series expansions and the theory of quasi- periodic functions; we recommend the first volume of Sternberg’s book (1969). 5.10
Additional solved problems Problem 1 Determine a central force field in which a particle of mass m is allowed to describe the orbit r = r 0 e
, where r 0 > 0 is fixed, c is a non-zero constant and ϕ is the polar angle. Compute ϕ = ϕ(t) and r = r(t). Solution
Setting u = 1/r we have u = u 0 e −cϕ , where u 0 = 1/r
0 . From equation (5.26) it follows that d du V e 1 u = − c 2 L 2 z m u, from which by integration we obtain V (r) = −
2 z 2mr 2 (1 + c
2 ). We find ϕ(t) starting from the conservation of the angular momentum: mr 2 ˙ ϕ = L z , from which it follows that mr 2 0 e 2cϕ(t) ˙ ϕ(t) = L z . The last relation can be integrated by separation of variables. Problem 2 Prove that in a central force field with potential energy V (r) = −αe
−kr /r, where α and k are two positive constants, for sufficiently small values of the angular momentum there can exist a stable circular orbit. Solution The effective potential energy is given by V e
L 2 z 2mr 2 − α e −kr
r ,
5.10 Motion in a central field 209 and hence lim r →∞ V e (r) = 0, lim r →0+
V e (r) = + ∞. Differentiating once we find V e (r) = − L 2 z mr 3 + αk
e −kr
r + α
e −kr
r 2 . Circular orbits correspond to critical points of the effective potential energy. Thus we must study the equation re −kr
(1 + kr) = L 2 z mα . (5.109) The function f (x) = xe −x (1 + x), when x ∈ R + varies, has a unique critical point (an absolute maximum) for x = ( √ 5 + 1)/2. If we set M = f ( √ 5 + 1)/2 , we find that (5.109) has a solution if and only if L 2 z ≤ mαM/k. In addition, if L
2 z
and maximum of V e , and hence there exists a stable circular orbit, whose radius tends to zero if L 2 z → 0. Problem 3 (From Milnor 1983, pp. 353–65.) Consider Kepler’s problem and suppose that the angular momentum is non- zero. Prove that, as time t varies, the velocity vector v = ˙r(t) moves along a circle C lying in a plane P passing through the origin. This circle, and its orientation, uniquely determines the orbit r = r(t). The orbit is elliptic, hyperbolic or parabolic according to whether the origin is at the interior, at the exterior, or exactly on the circle C. Solution
Let R = k/L z . From equation (5.6) (with V (r) = −k/r) it follows that dv/dϕ = −Rr/r which yields, by integration, v = Re ϕ + c, with c = (c 1 , c
2 ) an integration constant. This shows that v moves along a circle C with centre c, radius R, lying in the same plane containing the orbit. Let = |c|/R be the distance of c from the origin, divided by the radius of the circle C. If we choose the orientation of the axes x, y of the plane so that c lies on the y-axis, we find v = R( − sin ϕ, + cos ϕ). (5.110) It follows that L z = mrR(1 +
cos ϕ) and
we find
again that
r = L z /[mR(1 + cos ϕ)], and hence equation (5.37) in which the eccentricity is given by . This shows how c lies at the interior or exterior, or it belongs to the circle C according to whether the orbit is elliptic, hyperbolic or parabolic. Problem 4 Prove that there are no equilibrium points for the n-body problem. 210 Motion in a central field 5.10 Solution
An equilibrium point is a solution of the system of equations −∇ r i V (r
1 , . . . , r n ) = 0,
i = 1, . . . , n, where V is the potential energy (see (5.103)). Since V is homogeneous of degree −1 we have − n i =1 r i · ∇
r i V = V. (5.111) However, since V is a sum of negative terms, it follows that V < 0, contrary to the requirement that, at an equilibrium point, the left-hand side of (5.111) vanishes. Problem 5 A central configuration in the n-body problem is a solution of (5.99) of the form r i
i , where ψ is a real function and the vectors a 1 , . . . , a n are
constant. Prove that if n = 3, for any values of the masses there exists a central configuration in which the three particles are at the vertices of an equilateral triangle (Lagrange’s solutions). Solution
Without loss of generality we can assume that the centre of mass of the system is fixed at the origin. Substituting r i = ψ(t)a
i into equation (5.98) we find |ψ| 3
−1 ¨ ψm i a i = j =i Gm i m j |a j − a i | 3 (a j − a i ). (5.112) Since the right-hand side is constant, by separation of variables we find ¨ ψ =
− λψ |ψ| 3 , (5.113) −λm i a i = j =i Gm i m j |a j − a
i | 3 (a j − a i ). (5.114) Equation (5.113) has infinitely many solutions while (5.114) can be written as ∇ r i V (a
1 , . . . , a n ) + λ
∇ r i I(a 1 , . . . , a n ) = 0.
(5.115) By the theorem of Lagrange multipliers, the system of vectors a 1 , . . . , a n yielding
a central configuration corresponds to an extremal point of the potential energy, under the constraint that the polar moment of inertia (see (5.104)) I is fixed. Let a
ij = |a i − a
j |. Then 1 ≤ i < j ≤ 3, M = 3 i
m i and since 4M I =
1≤i i m j a 2 ij , 5.10 Motion in a central field 211 equation (5.115) can be written as − Gm i m j a 2 ij + λm i m j a ij 2M = 0
(5.116) and admits the unique solution a ij = (2M G/λ) 1/3 . Problem 6 A limiting case of the three-body problem is the so-called reduced plane circular three-body problem, when one considers the motion of a point particle (of mass m) under the action of two masses moving along circular orbits. Suppose that the two masses are α and 1 − α, with centre of mass at the origin. Show that in the frame of reference moving with the two points, there exist three equilibrium configurations when the three bodies are aligned. Solution
The chosen reference system rotates with angular velocity ω. Choose the axes in such a way that the coordinates of the points of mass α and 1 − α are (1 − α, 0) and (
−α, 0), respectively. The kinetic energy of the point particle is T (x, y, ˙ x, ˙ y) =
1 2 m[ ˙ x 2 + ˙ y 2 + 2ωy ˙ x − 2ωx ˙y + ω 2 x
+ ω 2 y 2 ], (5.117) while the potential energy is V (x, y) = − Gmα
(x − 1 + α)
2 + y
2 − Gm(1 − α) (x + α)
2 + y
2 . (5.118) It follows that the equations of motion are m¨ x − 2mω ˙y = ∂V e ∂x (x, y),
m¨ y − 2mω ˙x = ∂V e ∂y (x, y), (5.119)
where V e (x, y) = V (x, y) − 1 2 mω 2 (x 2 + y
2 ). Hence the equilibrium positions are given by ˙ x = ˙
y = 0 and solutions (x, y) of ∇V e (x, y) = 0. Along the axis y = 0 we find
V e (x, 0) = 1 2 mω 2 x 2 ± Gmα
x − 1 + α
± Gm(1
− α) x + α
, (5.120)
where the signs are chosen in such a way that each term is positive: if x < −α the signs are ( −, −), if −α < x < 1 − α we have (−, +), if x > 1 − α we have (+, +). Since V e (x, 0)
→ +∞ when x → ±∞, x → −α, x → 1 − α and V e (x, 0) is 212 Motion in a central field 5.10 a convex function (verify that ∂ 2 V e ∂x 2 (x, 0) > 0 for all x), V e has exactly one critical point in each of these three intervals. Problem 7 An alternative way to solve Kepler’s equation (5.75) is obtained by computing the Fourier series expansion (see Appendix 7) of the eccentric anomaly as a function of the mean anomaly. Using the definition I n
1 2π 2π 0 cos(ny
− x sin y) dy (5.121)
of the nth Bessel function prove that ξ = l +
∞ n =1 2 n I n (ne) sin(nl). (5.122) Solution
Let e sin ξ = ∞ n =1 A n sin(nl), where the coefficients A n are given by A n = (2/π) π 0 e sin ξ sin(nl) dl. Integrating by parts and using the substitution l = ξ − e sin ξ, we find A n
2 nπ π 0 cos(nl)
d dl (e sin ξ) dl = 2 nπ π 0 cos(nl)
e cos ξ 1 − e cos ξ dl = 2 nπ π 0 cos(nl) dξ dl − 1 dl = 2 nπ π 0 cos[n(ξ − e sin ξ)] dξ = 2 n I n (ne). Recalling the power series expansion of the Bessel functions I n (x) = ∞ k =0 ( −1) k x n +2k 2 n +2k k!(n + k)! , (5.123)
it is possible to verify that (5.122) leads to (5.81)–(5.83) and vice versa. It is, however, necessary to appeal to equation (5.121) to get a better version of Theorem 5.4 and to prove the convergence of the series expansions (5.81)–(5.83) for all values of the eccentricity |e| < 0.6627434 . . .. The series (5.122) converges for all values of e ∈ [0, 1] (see Watson 1980, in particular Chapter XVII ‘Kapteyn series’ ). 6 RIGID BODIES: GEOMETRY AND KINEMATICS 6.1
Geometric properties. The Euler angles D efinition 6.1 A rigid body is a set of points {P 1 , . . . , P n }, n ≥ 2 which satisfy the rigidity constraints |P i − P j | = r ij > 0,
1 ≤ i < j ≤ n, (6.1) where r
ij are prescribed lengths. We note that: (a) the constraints (6.1) must satisfy certain compatibility conditions (e.g. the triangle inequalities r ij
ik + r
kj , 1
≤ i < k < j ≤ n); (b) the n(n − 1)/2 equations (6.1) are not all independent when n > 3. Regarding (b), if we consider three points P 1 , P
2 , P
3 in the system, not lying on the same line, for a chosen reference frame Σ = ( Ω , ξ
1 , ξ
2 , ξ
3 ) we can assign to P 1
2 two coordinates, and to the point P 3 one
coordinate. It is clear that having fixed the configuration of the triangle P 1 P
P 3 with respect to Σ , the coordinates of every other point in the system are automatically determined as functions of the six prescribed parameters. From this follows the well-known property stating that a rigid system containing at least three points, not all lying on the same line, has six degrees of freedom. In the case of a rigid body with all points lying on the same line, the number of degrees of freedom is five (it is sufficient to determine the configuration of the segment P 1 P
with respect to Σ ), while for a rigid body in the plane the degrees of freedom are three. We now introduce the concept of a body reference frame. Since the configuration of a rigid body is determined by that of any triangle P 1 P 2 P 3 formed by three of its points, we can associate with it a frame S ≡ (O, x 1
2 , x
3 ), called a body frame, for example assuming that the origin O coincides with P 1 , the x 1 -axis
contains the side P 1 P 2 , and the plane (x 1 , x
2 ) contains the triangle P 1 P
P 3 and the x 3 -axis, with the orientation of the unit vector e 3 = e
1 × e
2 , where e 1 and
e 2 are the unit vectors of the x 1 - and x
2 -axes (Fig. 6.1). Hence the problem of the determination of the configuration of a rigid body with respect to a frame Σ (which we call ‘fixed’) is equivalent to the problem of determining the configuration with respect to Σ of one of its body frames S. A body frame is determined by choosing an element g of the group of orientation-preserving isometries of three-dimensional Euclidean space (we exclude reflections). We denote by SO(3) the group of real 3 × 3 orientation-preserving 214 Rigid bodies: geometry and kinematics 6.1
3
1
2 V x 3
2
1
≡ P
2
3 Fig. 6.1 Body reference frame. orthogonal matrices A, i.e. such that AA T = 1 and det A = +1. Then g ∈ R
3 × SO(3) is the composition of a translation Σ →
, determined by the three coordinates of O in Σ , and of a rotation Σ → S around O. The column vectors of the rotation matrix A are the direction cosines of the axes of S with respect to Σ (nine parameters related by the six orthonormality conditions). This proves the following important property. P roposition 6.1 The configuration space of a rigid body with at least three non- collinear points in three-dimensional Euclidean space is R 3 × SO(3). If the system has a fixed point it must have three degrees of freedom and its configuration space is SO(3). A more direct representation of the transformation Σ → S is given by the so-called Euler angles (Fig. 6.2). It is easy to verify that the transformation A :
Σ → S can be obtained by composing three rotations. Let N be the line of nodes, i.e. the intersection between the planes x 3 = 0 and ξ 3 = 0 and denote by A a
A = A x 3 (ϕ)A N (θ)A ξ 3 (ψ), (6.2) 6.1 Rigid bodies: geometry and kinematics 215
Ј 3 j Ј 2 j Ј 1 x 3
1
2
O u w c Fig. 6.2
since (a) the rotation A ξ 3
3 makes the axis ξ 1
(b) the rotation A N (θ) by an angle θ (nutation) around the line of nodes N brings the axis ξ 3 to coincide with the axis x 3 ; (c) the rotation A x 3 (ϕ) by an angle ϕ around the axis x 3 makes the axes ξ 1 , ξ
3 coincide with the axes x 1 , x
2 . Remark 6.1 In order to determine the configurations of a rigid body it is not important to know the actual geometric structure of the system. We only need to know if the system has three non-collinear points. Often one considers rigid bodies subject to additional constraints. In general, given a set of N rigid bodies each with an associated vector q (i)
with six components (three in the planar case), we can impose constraints of the following kind: f j
(1) , . . . , q (N ) ) = 0,
j = 1, . . . , m, (6.3)
subject to the same criteria as seen in Chapter 1. Example 6.1 We study the planar system made up of two rods AB, CD of length 1 , 2 , respectively ( 1 ≥ 2 ), with the endpoints A, C fixed at a distance and with
the endpoint D constrained to lie on the segment AB (Fig. 6.3). 216 Rigid bodies: geometry and kinematics 6.2
Ј
Ј
1
2
Fig. 6.3
The constraint imposed on D makes sense only if < 1 + 2 . In this case the system has only one degree of freedom. We can take the angle α as a Lagrangian coordinate. We distinguish the following cases. (1) 2
/ 2 = ξ we have sin β = ξ sin α, γ = π − [α + β(α)], from which we can obtain the values of sin γ, cos γ, if we note that cos β = 1 − ξ 2 sin
2 α (β is always acute, as its opposite side is not the longest of the three sides, in any configuration). Since
dγ dα = −1 − dβ dα and dβ dα = ξ cos α
cos β , it is easy to check that dγ/dα is bounded and does not vanish, and hence the constraint is non-singular. (2)
2 < . Since now ξ > 1 we have the restriction |sin α| < 1/ξ. The extreme values correspond to β = ±π/2 and give a singularity of dγ/dα; this is the configuration which lies between those characterised by cos β = 1 − ξ 2 sin
2 α and those for which cos β = − 1 − ξ 2 sin 2 α. (3) 2 = . The configuration with D ≡ A is degenerate; it can be obtained as a limiting case for α → π/2, but if the point D remains fixed then the rod AB can rotate around A. 6.2 The kinematics of rigid bodies. The fundamental formula Since the configuration of a rigid system with respect to a frame Σ is determined by the configuration of any of its body frames S, the study of the kinematics of rigid systems consists of the description of the motion of S with respect to Σ .
Σ is the
following. T heorem 6.1 The velocity field of a rigid motion is expressed by the formula v(P ) = v(O) + ω × (P − O), (6.4)
6.2 Rigid bodies: geometry and kinematics 217 where v(O) is the velocity of a prescribed point O and ω is called the angular velocity. Equation (6.4) is a consequence of the following result. T heorem 6.2 In every rigid motion there exists one and only one vector ω, which is a function of time, through which we can express the variation of any unit vector e in the body by means of the formula de dt
ω × e (6.5)
(Poisson’s formula, also known as the attitude equation). Proof
The variation of a unit vector e in the time interval (t, t + ∆ t) can be expressed as e(t +
∆ t) − e(t) = A( ∆ t)e(t)
− e(t) = (A( ∆ t) − 1)e(t), where A(
∆ t) is an orthogonal matrix such that A(0) = 1. It follows that de dt
A(0)e(t). (6.6)
Recalling Lemma 1.1, which states that ˙ A(0) is a skew-symmetric matrix, if we write it in the form ˙ A(0) = Ω = ⎛ ⎝ 0 −ω 3 ω 2 ω 3 0 −ω 1 −ω 2 ω 1 0 ⎞ ⎠ , (6.7) where ω
1 , ω
2 , ω
3 define the vector ω(t), we see that (6.6) can be written exactly in the form (6.5), as Ω e =
ω × e. The uniqueness of the vector ω follows by noting that if there existed a different vector
ω with the same characterisation, the difference would satisfy ( ω − ω ) × e = 0 (6.8) for every unit vector e of S, implying ω = ω . It is interesting to note that if a unit vector e(t) satisfies equation (6.5) then it must be fixed in the body frame, as these are the only unit vectors satisfying the transformation law (6.6). The Poisson formula is clearly valid for any fixed vector in the body frame W: dW dt = ω × W.
(6.9) For any two points P, O in S, by applying equation (6.9) to the vector W = P −O, one finds the formula (6.4). 218 Rigid bodies: geometry and kinematics 6.2 Clearly the condition ω = 0 characterises any motion which is not purely a translation. P roposition 6.2 A necessary and sufficient condition for a rigid motion to be a translation is that ω = 0.
P roposition 6.3 If the direction of ω is constant in the reference frame Σ then
it is constant in the body frame S, and vice versa. Proof
Suppose that e = ω/|ω| is constant in S. We can then apply to it equation (6.5) and conclude that de/dt = 0, i.e. that e does not vary in Σ . Conversely, if de/dt = 0 in Σ , the unit vector e satisfies (6.5), since trivially ω × e = 0. It follows that e is constant in S. Rigid systems can be subject to non-holonomic constraints. This is illustrated in the following example. Example 6.2 Show that a disc that rolls without sliding on a plane π, and always orthogonal to it, is not a holonomic system (Fig. 6.4). If we ignored the kinematic condition which requires that the velocity at the contact point is zero, we would have a holonomic system with four degrees of freedom. The corresponding Lagrangian coordinates can be taken as: – two coordinates x 1 , x 2 which determine the position of the centre O; – the angle ψ formed by the plane of the disc with a fixed plane, normal to the base plane; – the angle ϕ formed by a radius fixed on the disc with the normal to the base plane.
O O w c Fig. 6.4
6.3 Rigid bodies: geometry and kinematics 219 If the addition of the kinematic constraint preserved the holonomic property of the system, then it would be possible to determine a relation of the kind f (x
1 , x
2 , ϕ) = 0; indeed this constraint cannot have any effect on the determination of the angle ψ. Hence to prove that the system is not holonomic, it is enough to prove that the coordinates x 1 , x
2 , ϕ remain independent. Hence we show that, for a given configuration of the system (for example, one associated with zero values for x 1 , x
2 , ϕ, ψ), we can move the disc in such a way that we can reach any other configuration, characterised by arbitrary values of x 1 , x 2 , ϕ, ψ. Ignoring the coordinate ψ, which can be chosen once the other three parameters are fixed, there exist infinitely many motions achieving this aim; if R is the radius of the given disc, and if x (0) 1
(0) 2 , ϕ (0) are the coordinates of the final configuration, it suffices to connect the points having coordinates (0, 0) and (x (0) 1 , x
(0) 2 ) in the plane π with an arc of any regular curve of length R(2kπ +ϕ (0) ), with k ≥ 0 any integer, allowing the disc to roll along this arc. This example shows that constraints that are not holonomic do not prevent certain configurations, but limit to certain classes the motions that connect two prescribed configurations (a familiar example is the parking of a car, when some manoeuvring is necessary due to the fact that the wheels cannot slide on the road).
Remark 6.2 If one prescribes a curve γ along which one constrains the disc of Example 6.2 to roll (x 1 = x 1 (q), x
2 = x
2 (q)) the system is again holonomic. Indeed, this gives a relation Rϕ =
q 0 (x 2 1 (η) + x 2 2 (η)) 1/2 dη + constant. If in addition the disc is constrained in the plane containing the unit vector tangent to γ then the system has only one degree of freedom, corresponding to the Lagrangian coordinate q. 6.3
Instantaneous axis of motion T heorem 6.3 If ω = 0 at a given time instant, then there exists at that instant a straight line parallel to ω, whose points have velocity parallel to ω or zero. Proof First of all, note that by taking the scalar product of (6.4) with ω we find v(P )
· ω = v(O) · ω, (6.10)
i.e. the product v · ω is invariant in the velocity field of a rigid system. We want to show that there exists a line, parallel to ω, along which the velocity reduces 220 Rigid bodies: geometry and kinematics 6.3 only to the component parallel to ω (common to all the points in the field). To this aim we consider the plane π normal to ω and passing through O and we look for P ∗ ∈ π with the property that v(P ∗ ) × ω = 0. Evaluating equation (6.4) at P ∈ π, and taking the vector product of both sides with ω, in view of the fact that ω · (P − O) = 0, we find v × ω = (P − O)ω 2 + v(O)
× ω. This yields the sought solution P ∗ as:
P ∗ − O = ω × v(O) ω 2 . (6.11)
In addition, from (6.4) we immediately deduce that all points belonging to the same line parallel to ω have the same velocity. This completes the proof of the theorem. D efinition 6.2 The straight line of Theorem 6.3 is called the instantaneous axis of motion (instantaneous axis of rotation, if the invariant v · ω vanishes). For the previous considerations it follows that the velocity field in a rigid motion has rotational symmetry with respect to the instantaneous axis of motion. Figure 6.5 justifies the fact that the generic rigid motion is called helical.
Download 10.87 Mb. Do'stlaringiz bilan baham: 
|